Number of Islands BFS and DFS in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BFS-based Number of Islands

What is the output of the following C++ code that counts islands using BFS?

DSA C++

int numIslands(vector<vector<char>>& grid) {
    int count = 0;
    int rows = grid.size();
    int cols = grid[0].size();
    vector<vector<bool>> visited(rows, vector<bool>(cols, false));
    vector<pair<int,int>> directions = {{1,0},{-1,0},{0,1},{0,-1}};
    for(int i=0; i<rows; i++) {
        for(int j=0; j<cols; j++) {
            if(grid[i][j] == '1' && !visited[i][j]) {
                count++;
                queue<pair<int,int>> q;
                q.push({i,j});
                visited[i][j] = true;
                while(!q.empty()) {
                    auto [r,c] = q.front(); q.pop();
                    for(auto& d : directions) {
                        int nr = r + d.first;
                        int nc = c + d.second;
                        if(nr>=0 && nr<rows && nc>=0 && nc<cols && grid[nr][nc]=='1' && !visited[nr][nc]) {
                            q.push({nr,nc});
                            visited[nr][nc] = true;
                        }
                    }
                }
            }
        }
    }
    return count;
}

int main() {
    vector<vector<char>> grid = {
        {'1','1','0','0','0'},
        {'1','1','0','0','0'},
        {'0','0','1','0','0'},
        {'0','0','0','1','1'}
    };
    cout << numIslands(grid) << endl;
    return 0;
}

A 3

B 2

C 1

D 4

💡 Hint

Count connected groups of '1's using BFS.

The grid has three separate islands: top-left block, middle single cell, and bottom-right block.

❓ Predict Output

intermediate

Output of DFS-based Number of Islands

What is the output of this C++ code that counts islands using DFS?

DSA C++

void dfs(int r, int c, vector<vector<char>>& grid, vector<vector<bool>>& visited) {
    int rows = grid.size();
    int cols = grid[0].size();
    if(r<0 || c<0 || r>=rows || c>=cols || grid[r][c]=='0' || visited[r][c]) return;
    visited[r][c] = true;
    dfs(r+1, c, grid, visited);
    dfs(r-1, c, grid, visited);
    dfs(r, c+1, grid, visited);
    dfs(r, c-1, grid, visited);
}

int numIslands(vector<vector<char>>& grid) {
    int count = 0;
    int rows = grid.size();
    int cols = grid[0].size();
    vector<vector<bool>> visited(rows, vector<bool>(cols, false));
    for(int i=0; i<rows; i++) {
        for(int j=0; j<cols; j++) {
            if(grid[i][j]=='1' && !visited[i][j]) {
                count++;
                dfs(i,j,grid,visited);
            }
        }
    }
    return count;
}

int main() {
    vector<vector<char>> grid = {
        {'1','1','0','0','0'},
        {'1','0','0','1','1'},
        {'0','0','1','0','0'},
        {'1','0','0','1','1'}
    };
    cout << numIslands(grid) << endl;
    return 0;
}

A 4

B 3

C 2

D 5

💡 Hint

DFS explores all connected land cells recursively.

There are five islands: top-left block, top-right block, middle single cell, bottom-left single cell, bottom-right block.

🔧 Debug

advanced

Identify the error in BFS island counting code

What error will this code cause when counting islands using BFS?

DSA C++

int numIslands(vector<vector<char>>& grid) {
    int count = 0;
    int rows = grid.size();
    int cols = grid[0].size();
    vector<vector<bool>> visited(rows, vector<bool>(cols, false));
    vector<pair<int,int>> directions = {{1,0},{-1,0},{0,1},{0,-1}};
    for(int i=0; i<=rows; i++) {
        for(int j=0; j<cols; j++) {
            if(grid[i][j] == '1' && !visited[i][j]) {
                count++;
                queue<pair<int,int>> q;
                q.push({i,j});
                visited[i][j] = true;
                while(!q.empty()) {
                    auto [r,c] = q.front(); q.pop();
                    for(auto& d : directions) {
                        int nr = r + d.first;
                        int nc = c + d.second;
                        if(nr>=0 && nr<rows && nc>=0 && nc<cols && grid[nr][nc]=='1' && !visited[nr][nc]) {
                            q.push({nr,nc});
                            visited[nr][nc] = true;
                        }
                    }
                }
            }
        }
    }
    return count;
}

A Runtime error due to out-of-bounds access on grid

B Logic error counting islands incorrectly

C Compilation error due to missing header

D No error, code runs correctly

💡 Hint

Check loop boundary conditions carefully.

The outer loop uses i <= rows which causes i to go out of bounds, leading to runtime error.

🧠 Conceptual

advanced

Why use BFS or DFS for Number of Islands?

Why are BFS and DFS suitable algorithms to count the number of islands in a grid?

A Because they sort the grid cells to group islands together

B Because they explore all connected land cells efficiently, marking visited cells to avoid recounting

C Because they use hashing to store island coordinates

D Because they convert the grid into a binary tree for easier traversal

💡 Hint

Think about how connected components are found in graphs.

BFS and DFS traverse all connected nodes (land cells) starting from one cell, marking them visited to count each island once.

🚀 Application

expert

Modify DFS to count islands diagonally connected

Given the DFS code for counting islands, which modification will correctly count islands connected diagonally as well?

DSA C++

void dfs(int r, int c, vector<vector<char>>& grid, vector<vector<bool>>& visited) {
    int rows = grid.size();
    int cols = grid[0].size();
    if(r<0 || c<0 || r>=rows || c>=cols || grid[r][c]=='0' || visited[r][c]) return;
    visited[r][c] = true;
    dfs(r+1, c, grid, visited);
    dfs(r-1, c, grid, visited);
    dfs(r, c+1, grid, visited);
    dfs(r, c-1, grid, visited);
}

// rest of code unchanged

A Use BFS instead of DFS to count diagonal islands

B Change grid to store diagonal neighbors explicitly

C Add calls to dfs(r+1, c+1), dfs(r+1, c-1), dfs(r-1, c+1), dfs(r-1, c-1) inside dfs function

D No change needed; current DFS counts diagonal connections

💡 Hint

Diagonal neighbors are cells at row±1 and col±1.

To count diagonal connections, DFS must explore all 8 neighbors, including diagonals.

Practice

1. What does the "Number of Islands" problem ask you to find in a 2D grid of '1's (land) and '0's (water)?

easy

A. Count total number of '1's in the grid

B. Count how many separate groups of connected '1's exist

C. Find the largest island size

D. Find the shortest path between two islands

Solution

1. Step 1: Understand the problem definition

   The problem asks to count distinct islands, where an island is a group of connected '1's horizontally or vertically.

2. Step 2: Clarify what is being counted

   We count how many separate clusters of '1's exist, not the total number of '1's or sizes.

3. Final Answer:

   Count how many separate groups of connected '1's exist -> Option B

4. Quick Check:

   Number of islands = count of connected '1' groups [OK]

Hint: Islands = groups of connected '1's, count groups not cells [OK]

Common Mistakes:

• Counting total '1's instead of groups
• Including diagonal connections as connected
• Confusing island size with island count

2. Which of the following C++ code snippets correctly initializes a 2D vector grid of chars with given rows and columns filled with '0'?

easy

A. std::vector grid(rows, std::vector<char>(cols, '0'));

B. std::vector<char> grid(rows, std::vector<char>(cols, '0'));

C. std::vector grid(cols, std::vector<char>(rows, '0'));

D. std::vector grid(rows, cols, '0');

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector of chars, use vector of vectors: outer vector size = rows, inner vector size = cols.

2. Step 2: Check each option

   std::vector grid(rows, std::vector<char>(cols, '0')); correctly creates 'rows' vectors each with 'cols' chars initialized to '0'. Others have wrong types or parameters.

3. Final Answer:

   std::vector grid(rows, std::vector<char>(cols, '0')); -> Option A

4. Quick Check:

   2D vector init = vector of vectors with sizes [OK]

Hint: Use vector(rows, vector<char>(cols, '0')) [OK]

Common Mistakes:

• Swapping rows and cols sizes
• Using wrong vector types
• Trying to initialize 2D vector with three parameters

3. Given the following grid and BFS-based code snippet to count islands, what is the output?

std::vector grid = {
  {'1','1','0','0'},
  {'1','0','0','1'},
  {'0','0','1','1'},
  {'0','0','0','0'}
};
int countIslands(const std::vector& grid) {
  int count = 0;
  int rows = grid.size();
  int cols = grid[0].size();
  std::vector visited(rows, std::vector<bool>(cols, false));
  std::queue> q;
  int directions[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
  for(int i=0; i=0 && nr=0 && nc

medium

A. 3

B. 4

C. 2

D. 1

Solution

1. Step 1: Identify islands in the grid

   Islands are groups of connected '1's. Here, islands are: top-left cluster (3 cells), single '1' at (1,3), and bottom cluster (2 cells at (2,2),(2,3)). Total 3 islands.

2. Step 2: Understand BFS traversal

   The BFS marks all connected '1's visited per island, so count increments once per island found.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Count islands = 3 [OK]

Hint: Count connected groups by BFS marking visited [OK]

Common Mistakes:

• Counting diagonal connections as islands
• Missing isolated '1's as separate islands
• Not marking visited correctly causing overcount

4. The following DFS code for counting islands has a bug. What is the bug?

void dfs(int r, int c, std::vector& grid) {
  int rows = grid.size();
  int cols = grid[0].size();
  if(r < 0 || c < 0 || r >= rows || c >= cols || grid[r][c] == '0')
    return;
  grid[r][c] = '0';
  dfs(r+1, c, grid);
  dfs(r-1, c, grid);
  dfs(r, c+1, grid);
  dfs(r, c-1, grid);
}

int numIslands(std::vector& grid) {
  int count = 0;
  for(int i=0; i

medium

A. The base case does not check for visited cells

B. The DFS does not check boundaries correctly

C. The DFS does not mark visited cells, causing infinite recursion

D. The grid is modified during DFS, which is incorrect

Solution

1. Step 1: Analyze DFS grid modification

   The DFS marks visited cells by changing '1' to '0' in the original grid, modifying input.

2. Step 2: Identify why this is a bug

   Modifying input grid may be unintended side effect; better to use a separate visited array to avoid altering input data.

3. Final Answer:

   The grid is modified during DFS, which is incorrect -> Option D

4. Quick Check:

   Modifying input grid = bug [OK]

Hint: Avoid changing input grid; use visited array instead [OK]

Common Mistakes:

• Ignoring that grid is changed in-place
• Assuming no visited tracking needed
• Misreading boundary checks as incorrect

5. You want to count islands in a grid where diagonal connections also count as connected (8 directions). How should you modify the BFS or DFS approach?

hard

A. Count islands by summing all '1's without traversal

B. Only check horizontal neighbors, ignore vertical and diagonal

C. Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array

D. Use BFS for horizontal and DFS for vertical connections separately

Solution

1. Step 1: Understand connectivity change

   Diagonal connections mean neighbors include 8 directions: up, down, left, right, and 4 diagonals.

2. Step 2: Modify directions array

   Add diagonal moves (1,1), (1,-1), (-1,1), (-1,-1) to directions used in BFS or DFS to explore all connected neighbors.

3. Final Answer:

   Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array -> Option C

4. Quick Check:

   Include all 8 neighbors for diagonal connectivity [OK]

Hint: Add diagonal moves to directions for 8-way connectivity [OK]

Common Mistakes:

• Ignoring diagonal neighbors
• Counting only horizontal or vertical neighbors
• Trying to count without traversal